2014-2015 ACM-ICPC, Asia Tokyo Regional Contest F題：There is No Alternative [最小生成树]

题意：N个点，M条边，问最小生成树的必须边的条数和权值和，必须边是去掉后最小生成树会扩大的边。

范围：N至多500，M至多10000

解法：必须边必然在某个最小生成树中，一开始直接求一遍最小生成树，然后枚举树中每条边去掉后的MST是否会变大即可。

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<iostream>
#include<stdlib.h>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<bitset>
#pragma comment(linker, "/STACK:1024000000,1024000000")
template <class T>
bool scanff(T &ret){ //Faster Input
char c; int sgn; T bit=0.1;
if(c=getchar(),c==EOF) return 0;
while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();
sgn=(c=='-')?-1:1;
ret=(c=='-')?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
if(c==' '||c=='\n'){ ret*=sgn; return 1; }
while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;
ret*=sgn;
return 1;
}
#define inf 1073741823
#define llinf 4611686018427387903LL
#define PI acos(-1.0)
#define lth (th<<1)
#define rth (th<<1|1)
#define rep(i,a,b) for(int i=int(a);i<=int(b);i++)
#define drep(i,a,b) for(int i=int(a);i>=int(b);i--)
#define gson(i,root) for(int i=ptx[root];~i;i=ed[i].next)
#define tdata int testnum;scanff(testnum);for(int cas=1;cas<=testnum;cas++)
#define mem(x,val) memset(x,val,sizeof(x))
#define mkp(a,b) make_pair(a,b)
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define pb(x) push_back(x)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

int n,m;
int eds1,eds2;
int b[111],a[111];
int q[111111];
int head,tail;
int dis[111111];
int t[111];
int swapx(int x,int y){
swap(t[x],t[y]);
int ret=0;
drep(i,n,1){
ret=ret<<1|t[i];
}
swap(t[x],t[y]);
return ret;
}
bool vis[111111];
void bfs(int root){
rep(i,1,33000)dis[i]=inf;
head=tail=0;
q[tail++]=root;
dis[root]=0;
while(head<tail){
int st=q[head++];
int x=st;
vis[x]=1;
rep(i,1,n){
t[i]=x&1;
x>>=1;
}
rep(i,1,n-1){
int nst;
nst=swapx(i,i+1);
if(!vis[nst]){
vis[nst]=1;
q[tail++]=nst;
dis[nst]=min(dis[st]+1,dis[nst]);
}
}
}
}

int c1[111],c2[111];
int main(){
scanff(n);
scanff(m);
rep(i,1,n)scanff(a[i]);
int st=0;
int x;
eds1=eds2=0;
int pd=0;
int tot=0;
rep(i,1,m){
pd^=1;
scanff(x);
rep(j,1,x){
tot++;
c1[tot]=pd;
c2[tot]=pd^1;
}
}
drep(i,n,1){
eds1=eds1<<1|c1[i];
eds2=eds2<<1|c2[i];
}
drep(i,n,1){
st=st<<1|a[i];
}
bfs(st);
printf("%d\n",min(dis[eds1],dis[eds2]));
return 0;
}